Question

Equation of tangent having slope 1 to the circle x^2+y^2-10x-8y+5=0 is

Answer 1

Answer:

There are 2 tangents with slope 1

   x-y + 6√2 - 1 = 0 and

   x-y - 6√2 - 1 = 0

Step-by-step explanation:

Equation of line with given slope m , and y-intercept c is given by

y = mx+c.

Given m =1, thus tangent would be in the form

y = x +c

For the given circle , centre is (5, 4) and radius is 6.

For any given line to be a tangent to any given circle, the perpendicular distance from the centre of the circle to tangent line should be equal to radius of the circle.

Thus, |5-4 +c| /√2 = 6, On cross multiplication, we get

|c + 1| = 6√2,

c + 1 = ±6√2,

c = 6√2-1 or -6√2-1.

Answer 2

✌️✌️ Hey mate,

➡️Answer:

There are 2 tangents with slope 1

   x-y + 6√2 - 1 = 0 and

   x-y - 6√2 - 1 = 0

Step-by-step explanation:

Equation of line with given slope m , and y-intercept c is given by

y = mx+c.

Given m =1, thus tangent would be in the form

y = x +c

For the given circle , centre is (5, 4) and radius is 6.

For any given line to be a tangent to any given circle, the perpendicular distance from the centre of the circle to tangent line should be equal to radius of the circle.

Thus, |5-4 +c| /√2 = 6, On cross multiplication, we get

|c + 1| = 6√2,

c + 1 = ±6√2,

c = 6√2-1 or -6√2-1.

Thanks..

Nice to help you ✌️✌️